A square planar complex is formed by hybridization :

The correct answer is Option (4) → \(s,\, \ p_x,\, \ p_y,\, \ d_{x^2 - y^2}\)

A square planar complex uses dsp² hybridisation. The four hybrid orbitals must point to the four corners of a square lying in one plane (the xy plane). The orbitals that lie in that plane are

• the s orbital (non-directional),

• p_x and p_y (along the x and y axes), and

• d_x²−y², whose lobes lie along the x and y axes.

Example: \( \ce{[Ni(CN)4]^{2-}} \), where the inner 3d_x²−y² orbital is used.

Why the other options are incorrect

• Option (1) \(s, p_x, p_y, p_z\) — Incorrect. sp³ hybridisation, which gives a tetrahedral, not square planar, shape.

• Option (2) \(s, p_x, p_y, d_{z^2}\) — Incorrect. d_z² points along the z axis, out of the plane of p_x and p_y.

• Option (3) \(s, p_x, p_z, d_{xy}\) — Incorrect. p_z is out of the xy plane, and the lobes of d_xy lie between the axes, not along them.

NCERT (2026-27 edition):
Valence Bond Theory – “… In the square planar complexes, the hybridisation involved is dsp². An example is [Ni(CN)₄]^2–.”